0
$\begingroup$

I know that the standard pre-emphasis function is $y[n] = x[n]-\alpha\cdot x[n-1]$ and $\alpha$ is often set as $0.97$.

However, when I was learning speech processing using Praat, the pre-emphasis factor in it is $6\ \rm dB/oct$ which means the signal emphasise $6\ \rm dB$ per octave increase. That's what confused me.

I've read lots of questions and answers, most of them are explaining how to calculate the value of $\alpha$ or the $\mathcal Z$-transformation. But What I'm confused is the connection between $\alpha$ and $6\ \rm dB/oct$.

I found this equation in the notes of Praat:

The pre-emphasis factor $\alpha$ is computed as: $$ \alpha = \exp(-2\pi F \Delta t) $$ where the $F$ above which the spectral slope will increase by $6\ \rm dB/oct$, $\Delta t$ is the sampling period of the sound

According to above notes, it seems that the $\alpha$ only influences the "beginning frequency" of pre-emphasis but not "how much it changes".

Hope I explained myself clearly enough.

$\endgroup$

1 Answer 1

2
$\begingroup$

You're right that the value $\alpha$ will not influence the slope of the filter (approx. $6$ dB/octave). That slope is determined by the filter order, which is $1$ in this case (1 delay element in the filter implementation). The value $\alpha$ only influences the filter's cut-off frequency.

$\endgroup$
6
  • $\begingroup$ what does y[n] and x[n] denotes in pre emphasis equation ? $\endgroup$ Commented Jun 14, 2020 at 10:00
  • $\begingroup$ @NaveenGabriel: $x[n]$ is the input sequence, and $y[n]$ is the output sequence. $\endgroup$
    – Matt L.
    Commented Jun 14, 2020 at 10:25
  • $\begingroup$ Yes, I know that but what is the input sequence ? Is this equation is evaluated in frequency domain? In particular..what is "n" in "x" ? $\endgroup$ Commented Jun 14, 2020 at 10:35
  • $\begingroup$ @NaveenGabriel: $x[n]$ and $y[n]$ are time domain sequences, $n$ is the time index. $\endgroup$
    – Matt L.
    Commented Jun 14, 2020 at 10:45
  • 1
    $\begingroup$ @NaveenGabriel: yes, that's basically it. $\endgroup$
    – Matt L.
    Commented Jun 14, 2020 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.